Let us show that \(d^2{\mathbb {E}}Z ^2 \mathbf {1}(dZ\leqslant n)=O(n)\) . Suppose the contrary: for some \(n_k\rightarrow \infty \)  (as \(k\rightarrow \infty \) ) and \(d_k=d(n_k)\) ,
\((d_k^2/n_k){\mathbb {E}}Z ^2 \mathbf {1}(d_kZ\leqslant n_k)\rightarrow \infty \) . If so, the right-hand side of the last inequality (with \(n_k,d_k\)  replacing \(n,d\) ) goes to infinity in probability. But this is impossible, because by Theorem 1 in [1]}, \({\mathbb {P}}( d_k(S_{n_k}/n_k-1)\leqslant d_k/n_k)={\mathbb {P}}( S_{n_k}\leqslant n_k+1)\geqslant 1/13\)  for all \(k\) . Thus, \(d^2{\mathbb {E}}Z ^2 \mathbf {1}(dZ\leqslant n)=O(n)\) .
